Timeline for Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation?
Current License: CC BY-SA 3.0
11 events
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| Mar 27, 2015 at 3:05 | vote | accept | Peter Crooks | ||
| Mar 26, 2015 at 21:14 | comment | added | Jim Humphreys | @Dave: Yes, I'm only considering simple Lie algebras. I've also tried to clarify my answer yet again but will give up edits at this point. | |
| Mar 26, 2015 at 21:13 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Mar 26, 2015 at 20:52 | comment | added | Dave Witte Morris | Jim, when you say the answer is always "yes", do you mean for the simple case? It's easy to cook up counterexamples for the non-simple case, because we just have to take make sure every weight of $\mathfrak{sl}_2$ occurs the same number of times in the two representations. | |
| Mar 26, 2015 at 18:46 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Mar 26, 2015 at 17:34 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Mar 25, 2015 at 22:54 | comment | added | Dave Witte Morris | When $\mathfrak{g}$ is simple, and we consider only root embeddings, I think your analysis is correct, except that we first have to eliminate cases where the representation corresponding to a long root is not isomorphic to the representation corresponding to a short root. Extrapolating from $B_2$, I would think they are usually not isomorphic. | |
| Mar 25, 2015 at 22:48 | comment | added | Jim Humphreys | @Dave: I was just having some second thoughts, so I edited my answer. I may still be oversimplifying. | |
| Mar 25, 2015 at 22:46 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Mar 25, 2015 at 22:44 | comment | added | Dave Witte Morris | Jim, the representations afforded by taking $\phi_1$ and $\phi_2$ to be root embeddings won't usually be isomorphic. For example, in type $B_2$, the adjoint representation for a short root decomposes into three isomorphic 3-dimensional irreducibles (plus a 1-dimensional trivial), but the adjoint representation for a long root has a 3-dimensional, two 2-dimensionals, and three 1-dimensionals. (Well, something like that, anyway.) | |
| Mar 25, 2015 at 22:27 | history | answered | Jim Humphreys | CC BY-SA 3.0 |